Limits of Polynomial Functions!

Level 2

Let $p(x)$ be a polynomial of degree $4$ having extremum at $x=1,2$ and $\lim_{x\rightarrow0}\left ( 1+\frac{p(x)}{x^2} \right )=2$ .

Determine the value of $p(2)$ .

1 solution

Atul Anand Sinha
Feb 5, 2014

Let $p(x)=ax^4+bx^3+cx^2+dx+e$

${p}'(1)={p}'(2)=0$

$\lim_{x\rightarrow 0}\left ( \frac{x^2+p(x)}{x^2} \right )=2$

$\implies p(0)=0 \implies d=0$

$\lim_{x\rightarrow 0}\left ( \frac{2x+{p}'(x)}{2x} \right )=2$

$\implies {p}'(0)=0\implies d=0$

$\lim_{x \rightarrow 0}\left ( \frac{2+{p}''(x)}{2} \right )=2$

$\implies c=1$

On solving, $a=1/4, b=-1$ .

So $p(x)=\frac{x^4}{4}-x^3+x^2$

$\implies p(2) = 0$ .

When p(0) = 0, it implies that e=0 .... I think by mistake you have written d=0

Durgesh Yadav - 5 years ago